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What happens to the discretization and the roundoff errors as the step size is decreased?

Short Answer

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Answer: Decreasing the step size reduces discretization error by providing better approximations of the continuous function and increasing the number of points representing it. However, smaller step sizes also result in more calculations, which can accumulate roundoff errors and potentially lead to larger overall errors. The optimal step size balances these two types of errors to produce the most accurate numerical solution possible.

Step by step solution

01

Understanding Discretization Error

Discretization error is the error introduced when we approximate a continuous function or process by a discrete set of points or intervals. For example, when solving a differential equation using a numerical method (like Euler's method), we divide the interval of interest into smaller steps and approximate the continuous function at discrete values. Discretization error is directly related to the step size, as smaller step sizes can lead to better approximations of the continuous function.
02

Understanding Roundoff Error

Roundoff error is the error caused by representing real numbers in a finite number of digits, which is necessary when performing calculations on a computer. When performing arithmetic operations (addition, subtraction, multiplication, and division), the roundoff error can accumulate, especially if the number of calculations is large. Smaller step sizes result in more calculations, which might cause roundoff errors to become significant.
03

Analyzing the effect of decreasing step size on discretization error

As the step size decreases, the approximation of the continuous function becomes better, since more points are used to represent it. Consequently, the discretization error will decrease with decreasing step size. In general, the relationship between the discretization error and step size can be modeled as: \[E_{discretization} \propto h^p\] where \(E_{discretization}\) is the discretization error, \(h\) is the step size, and \(p\) is the order of the numerical method (e.g., \(p = 1\) for Euler's method).
04

Analyzing the effect of decreasing step size on roundoff error

As we decrease the step size, the number of calculations required to represent the continuous function increases. As a result, the roundoff errors in each calculation can accumulate and potentially result in a larger overall error. The relationship between roundoff error and step size can be modeled as: \[E_{roundoff} \propto \frac{1}{h}\] where \(E_{roundoff}\) is the roundoff error and \(h\) is the step size.
05

Balancing Discretization and Roundoff Errors

When choosing an appropriate step size for a numerical method, it is essential to balance the effects of discretization error and roundoff error. Decreasing the step size will reduce the discretization error but may increase roundoff error. If the step size is too small, roundoff errors become dominant, and the accuracy of the numerics deteriorates. Conversely, if the step size is too large, discretization errors become dominant. In summary, as the step size decreases, discretization error decreases, and roundoff error generally increases. The optimal step size balances these two types of errors to produce the most accurate numerical solution possible.

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Most popular questions from this chapter

Consider a long concrete dam $(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\left.\alpha_{s}=0.7\right)$ of triangular cross section whose exposed surface is subjected to solar heat flux of $\dot{q}_{s}=800 \mathrm{~W} / \mathrm{m}^{2}$ and to convection and radiation to the environment at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The 2-m-high vertical section of the dam is subjected to convection by water at \(15^{\circ} \mathrm{C}\) with a heat transfer coefficient of $150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and heat transfer through the 2 -m-long base is considered to be negligible. Using the finite difference method with a mesh size of \(\Delta x=\Delta y=1 \mathrm{~m}\) and assuming steady two-dimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surface of the dam. Answers: \(21.3^{\circ} \mathrm{C}\). $43.2^{\circ} \mathrm{C}, 43.6^{\circ} \mathrm{C}$

Consider transient heat conduction in a plane wall whose left surface (node 0 ) is maintained at \(80^{\circ} \mathrm{C}\) while the right surface (node 6) is subjected to a solar heat flux of \(600 \mathrm{~W} / \mathrm{m}^{2}\). The wall is initially at a uniform temperature of \(50^{\circ} \mathrm{C}\). Express the explicit finite difference formulation of the boundary nodes 0 and 6 for the case of no heat generation. Also, obtain the finite difference formulation for the total amount of heat transfer at the left boundary during the first three time steps.

Starting with an energy balance on a volume element, obtain the two- dimensional transient implicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, t)\) for the case of constant thermal conductivity and no heat generation.

Quench hardening is a mechanical process in which the ferrous metals or alloys are first heated and then quickly cooled down to improve their physical properties and avoid phase transformation. Consider a $40-\mathrm{cm} \times 20-\mathrm{cm}\( block of copper alloy \)\left(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ being heated uniformly until it reaches a temperature of \(800^{\circ} \mathrm{C}\). It is then suddenly immersed into the water bath maintained at \(15^{\circ} \mathrm{C}\) with $h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ for quenching process. However, the upper surface of the metal is not submerged in the water and is exposed to air at $15^{\circ} \mathrm{C}\( with a convective heat transfer coefficient of \)10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Using an explicit finite difference formulation, calculate the temperature distribution in the copper alloy block after 10 min have elapsed using \(\Delta t=10 \mathrm{~s}\) and a uniform mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\).

Consider transient heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0 , \(1,2,3\), and 4 with a uniform nodal spacing of $\Delta x$. The wall is initially at a specified temperature. The temperature at the right boundary (node 4) is specified. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary node 0 for the case of combined convection, radiation, and heat flux at the left boundary with an emissivity of \(\varepsilon\), convection coefficient of \(h\), ambient temperature of \(T_{\infty}\), surrounding temperature of \(T_{\text {surr }}\), and uniform heat flux of \(\dot{q}_{0}\) toward the wall. Also, obtain the finite difference formulation for the total amount of heat transfer at the right boundary for the first 20 time steps.

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